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1 year ago

Combine terms/simolify

Mathematics

1 answer:

-Dominant- [34]1 year ago

5 0

Answer:

$\dfrac{2m -1}{5}$

Step-by-step explanation:

Since we have common denominators inside the parentheses, we can combine the denominators. Then, we get the following expression:

$2\huge\text{(}\dfrac{1}{5} m - \dfrac{2}{5} \huge\text{)} + \dfrac{3}{5}$

$2\huge\text{(}\dfrac{m - 2}{5} \huge\text{)} + \dfrac{3}{5}$

Now, we can multiply the expression inside the parentheses by 2.

$\huge\text{(}\dfrac{2m - 4}{5} \huge\text{)} + \dfrac{3}{5}$

Since the expression inside the parentheses cannot be <u>simplified further</u>, we can open the parentheses and simplify the expression.

$\dfrac{2m - 4}{5} + \dfrac{3}{5}$

We can combine the denominators as the denominators are the same.

$\dfrac{2m - 4 + 3}{5}$

Finally, we can simplify -4 + 3.

$\dfrac{2m -1}{5}$

Therefore, the simplified expression is (2m - 1)/5.

Learn more about combining like terms: brainly.com/question/13080103

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Suppose that 5 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 51 cm. (a) How much work

olganol [36]

Answer:

a) 0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The spring must be 5.6 centimeters far from its natural length.

Step-by-step explanation:

a) The work done to stretch the ideal spring from its natural length is defined by the following definition:

$W = \frac{1}{2}\cdot k\cdot (x_{f}-x_{o})^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons.

$x_{o}$ , $x_{f}$ - Initial and final lengths of the spring, measured in meters.

$W$ - Work, measured in joules.

The spring constant is: ( $W = 5\,J$ , $x_{o} = 0.36\,m$ , $x_{f} = 0.51\,m$ )

$k = \frac{2\cdot W}{(x_{f}-x_{o})^{2}}$

$k = \frac{2\cdot (5\,J)}{(0.51\,m-0.36\,m)^{2}}$

$k = 444.44\,\frac{N}{m}$

If we know that $k = 444.44\,\frac{N}{m}$ , $x_{o} = 0.41\,m$ and $x_{f} = 0.46\,m$ , then the work needed is:

$W = \frac{1}{2}\cdot \left(444.44\,\frac{N}{m} \right)\cdot (0.46\,m-0.41\,m)^{2}$

$W = 0.555\,J$

0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The elastic force of the ideal spring ( $F$ ), measured in newtons, is defined by the following formula:

$F = k\cdot \Delta x$ (2)

Where $\Delta x$ is the linear difference from natural length, measured in meters.

If we know that $k = 444.44\,\frac{N}{m}$ and $F = 25\,N$ , then the linear difference is:

$\Delta x = \frac{F}{k}$

$\Delta x = \frac{25\,N}{444.44\,\frac{N}{m} }$

$\Delta x = 0.056\,m$

The spring must be 5.6 centimeters far from its natural length.

3 0

3 years ago

Please explain if is proportional or not

ryzh [129]

Answer:

Yes

Step-by-step explanation:

1x6=6

2x6=12

3x6=18

4x6=24

4 0

2 years ago

Read 2 more answers

My mother pays my brother twice as much allowance as she pays me because he cleans the bathrooms. Together we get $24 each week.

Harman [31]

Answer:

Your brother gets $16 and you get $8.

Step-by-step explanation:

first of all what is 16+8=24

first step..i started from 2 and did twice of that wich is 4 but that only got $6

second step.... just keep doing your multiples....

third step....as soon as i got to the number 8 i did twice of that and i got 16 then i added those and i got $24

and that is your answer.

4 0

3 years ago

The weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 l

joja [24]

1. Know what you're looking for:

The range is the difference between the lowest and highest values.

Your weight is 145.2 and can vary from the actual weight by maximum 0.3 lbs less or maximum 0.3 lbs more.

2. Calculate your lowest and highest values :

Lowest : 145.2 - 0.3 = 144.9 lbs

Highest : 145.2 + 0.3 = 145.5 lbs

3. Calculate the range of actual weights of the object :

145.5 - 144.9 = 0.6 lbs

--> The answer is: <u>The range of actual weights of the object is 0.6 lbs.</u>

There you go! I really hope this helped, if there's anything just let me know! :)

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3 years ago

4) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is

lisov135 [29]

Answer:

l=0.1401P\\

w =0.2801P

where P = perimeter

Step-by-step explanation:

Given that a window is in the form of a rectangle surmounted by a semicircle.

Perimeter of window =2l+\pid/2+w

$P= 2l+3.14 (w/2)+w$

Or $P = 2l+2.57w\\l = \frac{P-2.57w}{2}$

To allow maximum light we must have maximum area

Area = area of rectangle + area of semi circle where rectangle width = diameter of semi circle

$A=lw +\pi \frac{w^2}{8}$

$A=lw +\pi \frac{w^2}{8}\\A=w*\frac{P-2.57w}{2}+0.3925w^2\\2A= Pw-2.57w^2+(0.785w^2)\\2A' = P-5.14w+1.57 w\\2A" =-5.14+1.57$

Hence we get maximum area when i derivative is 0

i.e. $P-5.14w+1.57 w=0\\3.57w =P\\w = \frac{P}{3.57} =0.2801P$

$l = \frac{P-2.57w}{2}\\l = 0.1401P$

Dimensions can be

$l=0.1401P\\w =0.2801P$

5 0

2 years ago